A natural gas may be assumed to be a mixture of methane and ethane only. Complete combustion of 10 L of gas at S.T.P. the heat evolved was 474.6 KJ. Assuming ${\left({\mathrm{\Delta }}_{\mathrm{C}}\mathrm{H}\right)}_{{\mathrm{CH}}_4\left(\mathrm{g}\right)}=-894\mathrm{KJ}.{\mathrm{mol}}^{-1}\mathrm{and}{\left({\mathrm{\Delta }}_{\mathrm{C}}\mathrm{H}\right)}_{{\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)}=-1500\mathrm{KJ}.{\mathrm{mol}}^{-1}$ .Calculate the percentage composition of the mixture by volume is around

Detailed Solution: Let us consider the combustion processes of methane and ethane.

$\begin{array}{l}{\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{C}}_2{\mathrm{H}}_6+\frac{7}{2}{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}\end{array}$

Let us consider the following

$\begin{array}{l}{n}_{\text{Total }}=\frac{10}{22.41}=0.44 \mathrm{mol};n_{{\mathrm{CH}}_4}=n_1;n_{C_2{H}_6}=n_2;n_1+n_2=0.44\\ -894\times n_1+(-1500)\times n_2=-474.6\\ n_1+1.68n_2=0.53 \text{ (2) }\\ \text{ (2) - (1) }\\ 0.68n_2=0.53-0.44\\ n_2=0.133\end{array}$

The percentage of methane in the mixture is $\frac{0.31}{0.44}\times 100=70.45$

And the remaining is ethane.